# How many different binary numbers can be formed using 8 bits if:

So the question says

How many different binary numbers can be formed using 8 bits if: 1- In each number there are 3 adjacent ones 2- In each number there are exactly 6 ones (Edit: Each Part is different and solved separately)

I solved easily using logic but the thing is, we need to solve using permutation and combination rules I don't have any clue about the right answers

I got 6 for the first and 28 for the 2nd

• Oh, read those as two conditions of one problem. Aug 5 at 23:25
• Your answer for $1$ implies the condition is that there are no other $1$s other than the three? Aug 5 at 23:27
• What is the problem composer's intent, with respect to the first problem? Is 1-1-1-1-0-0-0-0 a satisfying sequence? What about 1-1-1-0-1-0-0-0? Aug 6 at 0:45
• For part $(2)$ there are $\displaystyle \binom{8}{2} = 28~$ ways of selecting two positions out of the $(8)$ for the $0$'s to be placed. This assumes sampling without replacement, where the order of selection is deemed irrelevant. Aug 6 at 0:47
• Oh I am sorry for the misunderstanding Each part is separated, and solved separately Aug 6 at 2:46